\(\angle B=180-\angle А- \angleС=180-\htmlClass{kA}{50}-\htmlClass{kC}{50}=\htmlClass{kB}{80}\degree\)
\(AB = {AC * sin(\angle C)\over sin(\angle B)} ={\htmlClass{kAC}{8} * sin( \htmlClass{kC}{50})\over sin( \htmlClass{kB}{80})}=\htmlClass{kAB}{6.223}\)
\(BC = {AC * sin(\angle A)\over sin(\angle B)} ={\htmlClass{kAC}{8} * sin( \htmlClass{kA}{50})\over sin( \htmlClass{kB}{80})}=\htmlClass{kBC}{6.223}\)
\(P=AB+BC+AC=\htmlClass{kAB}{6.223}+\htmlClass{kBC}{6.223}+\htmlClass{kAC}{8}=\htmlClass{kP}{20.446}\)
\(S=\htmlClass{fontm}{ \sqrt{{P\over 2}*({P\over 2}-AC)*({P\over 2}-BC)*({P\over 2}-AB)}}=\htmlClass{fontm}{ \sqrt{{\htmlClass{kP}{20.446}\over 2}*({\htmlClass{kP}{20.446}\over 2}-\htmlClass{kAC}{8})*({\htmlClass{kP}{20.446}\over 2}-\htmlClass{kBC}{6.223})*({\htmlClass{kP}{20.446}\over 2}-\htmlClass{kAB}{6.223})}}=\htmlClass{kS}{19.068}\)
\(H={2 * S\over AC}={2 * \htmlClass{kS}{19.068}\over \htmlClass{kAC}{8}}=\htmlClass{kH}{4.767}\)
\( Дано: \triangle ABC;\)
\( AC=\htmlClass{kAC}{8} см;\)
\( \angle A=\htmlClass{kA}{50} \degree;\)
\( \angle C=\htmlClass{kC}{50} \degree;\)
\( Найти: P _{\triangle ABC} ?\)
\( Решение:\)
\(P=AB+BC+AC\)
\(\angle B=180-\angle А- \angleС=180-\htmlClass{kA}{50}-\htmlClass{kC}{50}=\htmlClass{kB}{80}\degree\)
\(AB = {AC * sin(\angle C)\over sin(\angle B)} ={\htmlClass{kAC}{8} * sin( \htmlClass{kC}{50})\over sin( \htmlClass{kB}{80})}=\htmlClass{kAB}{6.223}см\)
\(BC = {AC * sin(\angle A)\over sin(\angle B)} ={\htmlClass{kAC}{8} * sin( \htmlClass{kA}{50})\over sin( \htmlClass{kB}{80})}=\htmlClass{kBC}{6.223}см\)
\(P=\htmlClass{kAB}{6.223}см+\htmlClass{kBC}{6.223}см+\htmlClass{kAC}{8}см=\htmlClass{kP}{20.446}см\)
\( Ответ:\htmlClass{kP}{20.446}см\)